Cube Roots of Unity and the Equation
The question asks for the solutions (\(z\)) to the equation \((z-100)^3 + 1000 = 0\). We are given that \(\omega \ne 1\) is a cube root of unity.
Key properties of the cube roots of unity (\(1, \omega, \omega^2\)) are:
- The cube roots satisfy the equation \(x^3 = 1\).
- They follow the rules: \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).
The value \(\omega\) represents one of the complex cube roots of unity, often \(e^{i 2\pi/3}\) or \(e^{i 4\pi/3}\).
Solving the Cubic Equation
First, we rearrange the given equation to isolate the cubic term:
\((z-100)^3 + 1000 = 0\)
Subtract 1000 from both sides:
\((z-100)^3 = -1000\)
To make the equation easier to work with, let's use a substitution. Let \(X = z - 100\). The equation now becomes:
\(X^3 = -1000\)
Finding the Cube Roots of -1000
We need to find the three values of \(X\) that satisfy \(X^3 = -1000\). We know that \((-10)^3 = -1000\), so one solution for \(X\) is \(X = -10\).
The three cube roots of a number \(C\) can be expressed using the principal cube root and the cube roots of unity. If \(\sqrt[3]{C}\) is the principal cube root, the three roots are \(\sqrt[3]{C}\), \(\sqrt[3]{C}\omega\), and \(\sqrt[3]{C}\omega^2\).
In this case, \(C = -1000\), and the principal cube root is \(-10\). Therefore, the three possible values for \(X\) are:
- \(X_1 = -10\)
- \(X_2 = -10 \times \omega = -10\omega\)
- \(X_3 = -10 \times \omega^2 = -10\omega^2\)
Deriving the Solutions for z
Now, we need to find the values of \(z\) by substituting back \(X = z - 100\) for each of the three possible values of \(X\) found above.
-
Case 1: \(X_1 = -10\)
\(z - 100 = -10\)
To find \(z\), add 100 to both sides:
\(z = 100 - 10\)
\(z = 90\)
-
Case 2: \(X_2 = -10\omega\)
\(z - 100 = -10\omega\)
Add 100 to both sides:
\(z = 100 - 10\omega\)
We can factor out 10 from the terms on the right side:
\(z = 10(10 - \omega)\)
-
Case 3: \(X_3 = -10\omega^2\)
\(z - 100 = -10\omega^2\)
Add 100 to both sides:
\(z = 100 - 10\omega^2\)
Factor out 10:
\(z = 10(10 - \omega^2)\)
Combining the results from the three cases, the solutions (\(z\)) for the equation \((z-100)^3 + 1000 = 0\) are \(90\), \(10(10 - \omega)\), and \(10(10 - \omega^2)\).
Similar Questions
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Important Questions from Complex Numbers
If \(x + iy = \sqrt {\frac{{a + ib}}{{c + id}}}\), then the value of x2 + y2 is -
If x = cos θ + i sin θ, then the value of \({x^n} + \frac{1}{{{x^n}}}\) is:
If \(\left| {\begin{array}{*{20}{c}} {6i}&{ - 3i}&1\\ 4&{3i}&{ - 1}\\ {20}&3&i \end{array}} \right| = x + iy\), then the values of x and y are:
If iz3 + z2 - z + i = 0, then the value of |z| is:
Nature of the triangle formed by the points representing the complex numbers 3 + 4i, 8 - 6i and 13 + 9i is:
